Question

$\large\red{\sf{Question}}$
If x, y and z are positive real numbers and a, b and c are rational numbers, then the value of

$\large\underline{\sf{ \frac{1}{1 + {x}^{b - a} + {x}^{ c- a} } + \frac{1}{1 + {x}^{a - b} {x}^{c - b} } + \frac{1}{1 + {x}^{ b- c} {x}^{ a- c} } }}$
​

Answer 1

Answer:

1

Step-by-step explanation:

$\frac{1}{1 + {x}^{b - a} + {x}^{c - a} } + \frac{1}{1 + {x}^{a - b} + {x}^{c - b} } + \frac{1}{1 + {x}^{a - c} + {x}^{b - c} }$

$\frac{1}{1 + \frac{ {x}^{b}}{{x}^{c} } + \frac{{x}^{c} }{ {x}^{a} } } + \frac{1}{1 + \frac{{x}^{a}}{ {x}^{b} } + \frac{{x}^{c}}{ {x}^{b} }} + \frac{1}{1 + \frac{{x}^{a}}{ {x}^{c} } + \frac{ {x}^{b}}{ {x}^{c} } }$

$\frac{ {x}^{a} }{ {x}^{a} +{x}^{b}+{x}^{c}} + \frac{ {x}^{b} }{ {x}^{b} +{x}^{a}+{x}^{c}} + \frac{ {x}^{c} }{ {x}^{c} +{x}^{a}+ {x}^{b}}$

$\frac{{ {x}^{a} +{x}^{b}+{x}^{c}}}{{ {x}^{a} +{x}^{b}+{x}^{c}}}$

1

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \dfrac{1}{1 + {x}^{b - a} + {x}^{ c- a} } + \dfrac{1}{1 + {x}^{a - b} {x}^{c - b} } + \dfrac{1}{1 + {x}^{ b- c} {x}^{ a- c}}$

We know, from laws of exponents,

$\boxed{ \rm{ \: {x}^{m - n} = \frac{ {x}^{m} }{ {x}^{n} } \: }}$

So, using this, the abobe expression can be rewritten as

$\rm = \dfrac{1}{1 + \dfrac{ {x}^{b} }{ {x}^{a} } + \dfrac{ {x}^{c} }{ {x}^{a} } } + \dfrac{1}{1 + \dfrac{ {x}^{c} }{ {x}^{b} } + \dfrac{ {x}^{a} }{ {x}^{b} } } + \dfrac{1}{1 + \dfrac{ {x}^{a} }{ {x}^{c} } + \dfrac{ {x}^{b} }{ {x}^{c} } }$

can be further rewritten as

$\rm = \dfrac{1}{\dfrac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} } } + \dfrac{1}{ \dfrac{ {x}^{b} + {x}^{c} + {x}^{a} }{ {x}^{b} } } + \dfrac{1}{\dfrac{ {x}^{c} + {x}^{a} + {x}^{b} }{ {x}^{c} } }$

can be rewritten as

$\rm \: = \: \dfrac{ {x}^{a} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \dfrac{ {x}^{b} }{ {x}^{a} + {x}^{b} + {x}^{c} } + \dfrac{ {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} }$

$\rm \: = \: \dfrac{ {x}^{a} + {x}^{b} + {x}^{c} }{ {x}^{a} + {x}^{b} + {x}^{c} }$

$\rm \: = \: 1$

Hence,

$\boxed{ \rm{ \:\dfrac{1}{1 + {x}^{b - a} + {x}^{ c- a} } + \dfrac{1}{1 + {x}^{a - b} {x}^{c - b} } + \dfrac{1}{1 + {x}^{ b- c} {x}^{ a- c}} = 1}}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ {x}^{0} = 1}\\ \\ \bigstar \: \bf{ {x}^{m} \times {x}^{n} = {x}^{m + n} }\\ \\ \bigstar \: \bf{ {( {x}^{m})}^{n} = {x}^{mn} }\\ \\\bigstar \: \bf{ {x}^{m} \div {x}^{n} = {x}^{m - n} }\\ \\ \bigstar \: \bf{ {x}^{ - n} = \dfrac{1}{ {x}^{n} } }\\ \\\bigstar \: \bf{ {\bigg(\dfrac{a}{b} \bigg) }^{ - n} = {\bigg(\dfrac{b}{a} \bigg) }^{n} }\\ \\\bigstar \: \bf{ {x}^{m} = {x}^{n}\rm\implies \:m = n }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$